Find y as a function of x if y"' + 9y' = 0, y(0) = 6, y'(0) = 9, y"(0) = 18. y(x) =

y(x) = 8 - 2cos 3x + 3sin 3x

Step-by-step explanation:

Given the equation:

y''' + 9y' = 0

The characteristic equation :

⇒r³ + 9r =0

Solving for r, we get r = 0 and r = ± 3i

The general equation for such equation is :

y = C₁ + C₂cos 3x + C₃sin 3x

Given:

y(0) = 6

Thus, Applying in the above equation, we get

6 = C₁ + C₂ 1

Differentiating y , we get:

y' = -3C₂sin 3x + 3C₃cos 3x

Given:

y'(0) = 9

Thus, Applying in the above equation, we get

9 = 3C₃

or,

C₃ = 3

Differentiating y' , we get:

y'' = -9C₂cos 3x - 9C₃sin 3x

Given:

y''(0) = 18

Thus, Applying in the above equation, we get

18 = -9C₂

or,

C₂ = -2

Applying in equation 1 , we get:

C₁ = 8

Thus,

y(x) = 8 - 2cos 3x + 3sin 3x

the answer is:

the answer is:

to solve the problem, we need to remember the quotient of power property, it's defined by the following relation:

if we have a quotienf of powers that have the same base, we need to keep the same base and subtract the exponent of the denominator power to the exponent of the numerator power.

so, we are given the expression:

then, calculating we have:

hence, the answer is:

have a nice day!